Mathematics for Engineering III

Transcription

1 Mathematics for Egieerig III Norto Uiversit Year

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3 Mathematics for Egieerig III

4 Cotets Page Chapter Partial Differetiatio Fuctios of Two or More Variables... Limits ad Cotiuit... Partial Derivatives... 4 Directioal Derivatives ad Gradiets Directioal Derivatives ad Gradiets of Two-Variable Fuct Directioal Derivatives ad Gradiets of Three-Variable Fuctio Total Differetial Chai ule... Chapter Eercises... Chapter Multiple Itegral Double Itegral...5. Defiitio...5. Properties of double itegrals...5. The Computatio of Double Itegral Chage of Variables i Double Itegrals...7 Triple Itegral...7. Defiitio...8. The Computatio of Triple Itegral...8. A z-simple egio Chage of Variables i Triple Itegrals...9 Applicatios.... Computatio of a Plae Area.... The Volume of a Solid.... Surface Area as a Double Itegral....4 Mass ad Ceter of Mass....5 Momets of Iertia...4 Chapter Eercises...5 Chapter Ordiar Differetial Equatio I Itroductio.... What is a differetial Equatio?.... Order of a Differetial Equatio.... Liear ad Noliear Differetial Equatios....4 Solutios....5 Implicit/Eplicit Solutio....6 Iitial-Value Problem (IVP)....7 Geeral Solutio of a Differetial Equatio...4 Separable Equatios...5 First Order Liear Equatios Beroulli Equatios iccati Equatio Eact Equatios Homogeeous Equatios...45

5 Chapter 4 Ordiar Differetial Equatio II Secod Order Noliear Equatios...49 Homogeeous Liear Differetial Equatio...5. Liear Idepedece...5. Wroskia...5 eductio of Order Homogeeous Liear Equatio with Costat Coefficiets Auiliar Equatio with Distict eal oots Auiliar Equatio with epeated eal oots No-homogeeous Liear Differetial Equatio with costat Coefficiets Usig the Method of eductio of Order to Fid a Particular Solutio Method of udetermied coefficiets Variatio of Parameters...64 Chapter 5 Liear Sstem of ODEs...67 Bibliograph...7

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7 Partial Differetiatio Chapter Partial Differetiatio Fuctios of Two or More Variables Defiitio A fuctio of two real variables, ad is a rule that assigs a uique real umber f ( to, ) each poit (, i some set D of the -plae. A fuctio of three variables,,,ad zis a rule that assigs a uique real umber f ( zto,, ) each poit (, zi, ) some set D of three-dimesioal space. The set D i these defiitios is the domai of the fuctio; it is the set of poits at which the fuctio is defied. I geeral, a fuctio of real variables,,,,, is regarded as a rule that assigs a uique real umber f (,,, ) to each poit (,,, ) i some set of -dimesioal space. Eample f :, f, is a fuctio of variables. If ad, the the value of the fuctio is f, 6. Note We ca deote z f (,,..., ) ad we call z the depedet variable ad,,..., the idepedet variables. For the fuctio of two variables z f (,, its domai is a set of poit (, ) -plae, o which f ( is, ) defied. The set of poit P(,, z f (, ) the graph of z f (,. It is a surface i -space. of the represets Eample State the domai of z f (, Solutio f is defied if +. Hece the domai of f is the poits o the disc with radius of uit. (Fig. ) Eample Fid the domai of z f (, Solutio Fig. Fig.

8 Partial Differetiatio > < z is defied if. Hece the domai is the set of poits > < iside of the rectagle. (See fig. ) Level Curves Each horizotal plae z C itersects the surface z f (, i a curve. The projectio of this curve o -plae is called a level curve. The surface z f (, Plae z c The level curve Limits ad Cotiuit Limit of a Fuctio of Two Variables The limit statemet lim f (, ) L or (, (, ) meas that for each give umber ε >, there eists a umber δ > so that wheever (, is a poit i the domai D of f such that the < + < δ f (, L < ε δ (, ) (, ) (, ) lim f, L ε >, δ > :, D, < + < δ f L < ε N.b: If the lim f (, ),, is ot the same for all approaches or paths withi the domai of f the the limit does ot eist. Eample + Evaluate lim (, ) (, ) Solutio + ( + ( For f (, ) +

9 Partial Differetiatio ( lim f, lim + (,,,, Eample If f (,, show that lim f + (, ) does t eist b evaluatig this limit,, alog the -ais, the -ais, ad alog the lie. Solutio First ote that the deomiator is zero at(, ), so f (,) is ot defied. If we approach the origi alog the -ais (where ), we fid that ( ) f (,) + So f (, as (, ) (, ) alog (ad ). If we approach the origi alog the -ais (where ), we fid that f (,) + So f (, as (, ) (, ) alog (ad ). However, alog the lie, the fuctioal values are f (, f (, ) for + so f (, as (, ) (, ) alog. Because f ( teds, ) toward differet umbers as,, alog the differet paths, it follows that f has o limit at the origi (, ). Eample Assumig each limit eists, evaluate: a. lim ( ),, (as: ) b. lim (, ) (, ) + Cotiuit of a Fuctio of Two Variables The fuctio f ( is, ) cotiuous at the poit(, ) if (i). f (, ) is defied. (ii). lim f (, eists. ),, (iii). lim f (, ) f(,,, (as: 4 5 ) The fuctio f is cotiuous o a set S if it is cotiuous at each poit i S. Limit ad Cotiuit for fuctio of three variables The limit statemet lim f z,, L ( z,, ) (,, z )

10 Partial Differetiatio meas that for each ε, δ f (,, ) > > such that (,, ) zis a poit i the domai of f such that f z L < ε wheever < ( ) + ( ) + ( z z) < δ The fuctio f ( zis,, ) cotiuous at the poit (,, ) (i). f (,, z ) (ii). lim f ( z,, ),,,, z z (iii). lim f ( z,, ) f(,, z),,,, z z P z if Partial Derivatives If z f (, the the partial derivatives of f with respect to ad are the fuctio f ad f, respectivel defied b f ( +Δ, f (, f (, lim Δ Δ f (, +Δ f(, ) f (, lim Δ Δ provided the limits eist. Eample f (, ) +. Fid f, f Solutio f, + f, + Alterative Notatios for Partial Derivatives For z f (,, the partial derivatives f, f are deoted b f z f(, ) f(, ) z D( f) f z f (, ) f(, ) z D( f) ab, are deoted The values of the partial derivatives of (, ) b f ( ab, ) ( f ab), ad f ( ab, ) f f at the poit ( ab, ) Eample Let z si + Solutio ( ). Evaluate z ( π,) 4

11 Partial Differetiatio z si( + ) + cos( + )( ) si( + ) + cos( + ) Thus, z π π siπ + cosπ ( π,) π π π ( ) + ( ) Eample Let f ( z,, ) + + z. Determie: f, f ad f z. Solutio f z,, + We treat, z as costats, the We treat, z as costats, the f ( z,, ) 4+ z We treat, as costats, the fz z,, z Eample Let z be defied implicitl as a fuctio of ad b the equatio z+ z Determie z ad z Solutio Differetiate implicitl with respect to, treatig as a costat: z z z + + z The solve this equatio for z : z z + z Similarl, holdig costat ad differetiatig implicitl with respect to, we fid z z + z + z So that z z + z Partial Derivative as a slope The lie parallel to the z-plae ad taget to the surface z f (, at the poit P(,, z ) has slope f (, ). Likewise, the taget lie to the surface at P that parallel to the z-plae has slope f (, ). Eample 4 Fid the slope of the lie that is parallel to the z-plae ad taget to the surface z + at the poit P (,, ) Solutio 5

12 Partial Differetiatio If f (, ) + ( + the the required slope is (, ) f (, ) f 4 +. Thus, (, ) f Note Higher-Order Partial Derivatives Give z f (,, the Secod order partial derivatives f f ( f ) f f f ( f f Mied secod-order partial derivatives f f ( f f f f ( f ) f The otatio f meas that we differetiate first with respect to ad the with respect to, while first ad the with respect to ). f meas just the opposite (differetiate with respect to Eample 5 For z f, 5 +, determie these higher-order partial derivatives. f f z a. b. c. d. f (, ) Solutio a. First differetiate with respect to, the to f + 9 f f ( + 9 ) b. Differetiate first with respect to ad the with respect to. f f f ( c. Differetiate with respect to twice: f f ( 6

13 Partial Differetiatio d. Evaluate the mied partial foud i part b at the poit (, ) f (, ). has mied secod-order partial ad f that are cotiuous i a ope disk cotaiig(, ), emark If the fuctio f (, ) derivatives f the (, ) (, ) f f I fact this remark is a theorem with the proof omitted here. Eample 6 Determie f, f, f, fwhere f (, ) e Solutio We have the partial derivatives f e f e + e The mied partial derivatives are f f e + e f e + e e ad f ( f ) e f f + e Eample 7 B direct calculatio, show that fz fz fz for the fuctio 4 f ( z,, ) z+ z. 4 Directioal Derivatives ad Gradiets 4. Directioal Derivatives ad Gradiets of Two-Variable Fuctio Directioal Derivative Let f be a fuctio of two variables, ad let u ui + uj be a uit vector. The directioal derivative of f at P (, ) i the directio of u is give b f ( + hu, + hu) f (, ) Df u (, ) lim h h provided the limit eists. Let f (, be a fuctio that is differetiable at P (, ). The f has a directioal derivative i the directio of the uit vector u u i+ u j give b D u f (, ) f(, ) u+ f(, ) u Eample Fid the derivative of f (, + at the poit P(, ) i the directio of the uit vector u i j Solutio The partial derivative f (, 4 ad f (,. The sice u ad u, we have 7

14 Partial Differetiatio Eample D u f (, ) f(, ) + f(, ) 6 The Gradiet Let f be a differetial fuctio at (, ad let f (, have partial derivatives f ( ad f (,. The the gradiet of f, deoted b f, is a vector, ) ) give b f f i f j (, ) (, ) + (, ) The value of the gradiet at the poit P (, ) (, ) + (, ) is deoted b f f i f j Fid the gradiet of the fuctio f (, + Solutio f (, ) ( + f (, ) ( + + the f (, ) i+ ( + j Theorem: If f is a differetiable fuctio of ad, the the directioal derivative of f at the poit P(, ) i the directio of the uit vector u is D u f (, ) f u (The proof is cosider a eercise) Eample Fid the directioal derivative f (, ) l( ) directio of v i j. Solutio f (, + + at the poit f 6, so (, ) 7 f (,, so f (, ) Thus, f f (, ) i j 6 6 A uit vector i the directio of v is v i j u ( i j) v + P, i the Thus 7 D (, f u u Directioal Derivatives ad Gradiets of Three-Variable Fuctio 8

15 Partial Differetiatio Directioal Derivatives Let f ( z,, ) be a differetiable fuctio at the poit P(,, z), ad let u ( u, u, u) be a uit vector. The directioal derivat ive of f at the poit P i the directio of u is give b D f,, z f,, z u + f,, z u + f,, z u u z Gradiet The gradiet of the the fuctio of three variable,, ad z f f zi,, + f z,, j+ f z,, k ad, hece, at a poit dirctio of a uit vector u is P(,, z) z u, the directioal derivative of f i the D f f z u (,, ) Eample 4 Fid the directioal derivative of f ( z,, ) z+ zat the poit (,, ) i the directio of the vector a i + j k. Solutio We ca fid f z,, f, z,, z, f z z Basic Properties of the gradiet Let f ad g be differetiable fuctios. The Costat rule: c for a costat c Liearit rule : af + bg a f + b g for costat a ad b Product rule : fg f g+ g f f g f f g Quotiet rule :, g g g Power rule : ( f ) f f (The proof are cosidered eercises) 5 The Total Differetial For a fuctio of oe v ariable d, f ( ), we defied the differetial d to be f d. For the two-variable case, we make the followig aalogous defiitio. Defiitio The total differetial of the fuctio f (, ) f f df d + d f(, d + f (, d where d ad d are idepedet variables. Similarl, for a fuctio of three w f,, z the total differetial is variables f f f df d + d + dz z is 9

16 Partial Differetiatio Eample Determie the total differetial of the give fuctios: a. f z + z Solutio 4,, 5 6 b. f (, ) l( ) f f f a. df z d d dz 6 d d 6dz 6 Chai ules The Chai rule for oe idepedet parameter Let f ( be, a differetiable fuctio of ad, ) ad let () t ad ( t) fuctios of t. The z f (, to be differetiable fuctio of t, ad dz z d z d + dt dt dt is a differetiable z d dt u z z u d dt Eample Le t z +, where ad t. Compute dz i two was: t dt a. b first epressig z eplicitl i terms of t. b. b usig the chai rule. Solutio a. B substitutig ad t, we fid that t z + + ( t ) t + t 4 for t t dz T hus, t + 4t dt b. Sie z + ad t, t, the z z d d,, t, t dt dt Use the chai rule for oe idepedet parameter: z z z u v u u u v u v dz z d z d + t + t t + 4t dt dt dt Etesios of the Chai ule Suppose z f (, is differetiable at (, ad that the partial derivatives of ( u, v) ad ( u, v) eist at ( u, v). The the composite fuctio z f [ ( u, v), ( u, v) ] is differetiable at ( u, v) with z z z z z z + ad + u u u v v v Eample

17 Partial Differetiatio Let z 4, where uv ad u v. Fid Solutio First fid the partial derivatives z ( 4 ) 4 ( uv ) v u u ( uv ) uv v v Therefore z z z + u u u z z ad u v ( 4 ) z u v u u u u v u v v 4v 4v + ( ( u v) 5 6u v EXECISES Fid the domai of the followig fuctio a z d z l ( b z ( + e + c z l ( + z + d z 4+ 4 Fid all first partial derivatives of each of fuctio f (, ) ( 4 f (, ( 4 ) f (, 4 f (, ) e cos 5 f (, ) esi 6 f (, ) 7 f (, ) e 8 f, arcta f, cos f r, θ r cosθ + f f Verif th at f, 5 f, + 5 f (, ) e cos f, arcta If F(,, fid F (, ) ad F (, ). F, l, fid, 4 ad F, 4. If F ( )

18 Partial Differetiatio f f A fuctio of two variables that satisfies Lapl ace s Equatio + is said to be harmoic. Show that the followig fuctios are harmoic fuctios. f, f, l f z,, z+ z fid each of the follo 5 If wig: a. f ( z,, ) b. f (,, ) c. f ( z,, ) (,, ) 4 6 If f z + + z, fid each of the followi g a. f ( z,, ) b. f (,, ) c. f zz ( z,, ) u u 7 The heat equatio c is the importat equatio i phsics (c is a costat). It is called t partial differetial equatio. Show that the fuctios ct ( 4ct) u e si ad u t e satisf the heat equatio. 8 Fid the idicated limit or state that it does ot eist. si ( + ) a. lim b. lim,, + + (, ) (, ) + lim d.,, lim c. 4 4,, + lim c. (, (,) ta ( + ) + lim 9 Show that does ot eit b cosiderig oe path to the origi alog the -ais,, + ad aother path alog th e lie. + Show that lim does t eist. (, (,) + Let f (, + 4 a. Show that f (, as (, ) (, ) b. Show that f (, as (, ) ( ) c. What coclusio do ou draw?. alog a straight lie m, alog the parabola. Supplemetar Eercise z z Show that the fuctio z ϕ ( + ) satisfies the equatio Fid the secod order partial derivatives of the fuctio z arcta Fid the total differetia fuctio of 4 Show that the fuctio f, l of the u satisfies Laplace equatio arcta u u +

19 Partial Differetiatio Fid the first order partial derivatives of the followig fuctios 5 z z z si, aswer: si, si 6 z z z, aswer:, l 7 z u u e + + aswer: + e + u z + + z u, e, + + ze z z 8 u + +z z arcsi + 9 Fid the total differetial of the followig fuctios z f (, + + si z l ( Fid z e +, f ad f (,) if f (, ) +. Aswer: f f 4 Let f (, ) e f, f, f, fid 5 Fid dz dt usig chai rule 4 a z, t, t b z l +, t, c z cos si, t, t z 6 Fid u ad z b the chai rule v a z 8 +, uv, u v b z ta, u v, u v c z, cos u, siv d z, u+ vlu, u vlv dz 7 Use chai rule to fid z, t, t+ 7 dt t f f 8 Use chai rule to fid the value of, u v (, ) +,, f u uv t u, v u, v if, 4,, 6 9 Use chai rule to fid the value of z r r, θ π 6 ad z θ r, θ π 6 Let r +, show that r r a b r r if z e, rcos θ, rsiθ. c r r r d r

20 Partial Differetiatio f Show that the followig fuctio satisfies Laplace equatio a, f esi + ecos b f (, ) l( + ) f, arcta Fid the gradiet f c f + a. f (, ) + f. f ( z,, ) + + z b. f (, ) g. (,, ) + + c. e h. f ( z,, ) e d. f (, ) cos e. f (, ) ( + Fid the gradiet vectors of the give fuctio at the give poit a. f, ;,) ( b. f ( (, + ;, c. f (, ;(, ) ) f z z z 4

21 Multiple Itegral Chapter Multiple Itegral Double Itegral z ( z f, (, ) k k. Defiitio Let be a regio i the -plae ad f (, ). I -plae, we draw the lies parallel to -ais ad -ais so that we get sub-rectagles i the regio. The sum of the area of each sub-rectagle is approimate to the area of the regio. A kth sub-rectagle whose area is defied b Δ A Δ Δ is the base of a solid with the altitude f (, k k) (, ) Vk f k k ΔA k ). The the volume of this kth solid is defied b Hece the volume of the solid whose base is the regio ad bouded above b the fuctio f ( is, approimate to k ( k, k) Ak V V f Δ k k This sum is called iema sums, ad the limit of the iema sums is deoted b f (, da lim f ( k, k ) + k ΔA which is called double itegral of z f (, k k over the regio.. Properties of double itegrals (i). Liearit rule: for costats a ad b af, + bg, da a f, da + b g, da Area ΔA k [ ] (ii). Domiace rule: if f ( g(,, throughout a regio, the 5

22 Multiple Itegral (, da g( da f, (iii). Subdivisio rule: If the regio is subdivided i two ad, the f (, da f (, da + f (, da. The Computatio of Double Itegral (i) Over ectagular egio If f (, is cotiuous over therectagle : a b, c d, the the double itegral f (, da ma be evaluated b either iterated itegral; that is, d b (, da f (, dd f (, dd f c a Eample compute ( da, where b d a c d c a [, ] [, ] ab cd is the rectagle with vertices (,), (,)(,, ) ad (, ). Aswer: 6 Eample 5 7 Evaluate da where is the rectagle,. Aswer: 8 (ii) Over Norectagular egios Tpe I regio This regio ca be described b the iequalities : a b, g g g ( ) b b g ( ) (, ) (, ) f da f dd a g Tpe II regio This regio ca be described b the iequalities : c d, h h d h ( (, ) (, ) f da f dd c h ( ) g a b d h c ( h ( 6

23 Multiple Itegral Eample Evaluate the double itegral ( ) + T da where T is the triagular regio eclosed b 4 the lies,,. Aswer:. Eample 4 Evaluate daover the regio eclosed betwee,,, 4. Aswer: 6.4 Chage of Variables i Double Itegrals Let Δ ad be the regios i -plae ( ) wher e Δ is the ew regio. A poit i, u, v, u, v z f, is this regio is defied b ( uv) where. If cotiuous over the regio, the f (, dd f (, ), (, ) u v u v J dudv where J is called Jacobia ad is defied b u v J u v Δ Special Case: Chage from Cartesia Coordiates to Polar Coordiates rcosθ rsiθ r θ the, M (, ) M ( r, θ ) J u u v v cosθ siθ r siθ r r cosθ Hece we obtai f (, dd f ( r cos θ, r siθ) rdrdθ Eample 5 Compute I + dd Eample 6, Δ a a Compute ( + ) is a regio defied b hemi circle +, >,. Aswer: 4 4 a π {(, ) :, } I dd, + Triple Itegral.Aswer: 4 π 7

24 Multiple Itegral. Defiitio z f z,, be a three- Let V be a solid ad variable fuctio def ied i. Ever plae parallel to each of the three coordiate plaes cut the solid V i to small parallelepipeds, sa, v k (see the figure) whose volume is defied b v k dddz. The the triple itegral of the fuctio d f ( z,, ) over the solid regio V is defied as follows: f (,, z) dddz lim f ( k, k, zk) Δvk V + k. The Computatio of Triple Itegrals Let V be a parallelepiped defied b the iequ alit a b, c d, m z the V db (,, ) (,, ) f zdv f zdddz m c a Eample Compute z e dv wherev (,, z) :,, z. V z dz V k { } As: e Eam ple Compute 8zdV where V [,] [, ] [,]. Aswer: V. A z-simple egio Suppose V is a solid regio that is bouded z f ad below above b the surface (, ) (, b the surface z f. The projectio of this solid regio o -plae results i regio i the plae. If w f (,, z) is a cotiuous fuctio over th e solid regio V, the we obtai V V d ( z ) M,, k k k z f, z f, f (, ) f (,, z) dv (,, z) dz da (, ) V f Eample Compute V dv where V is a solid i the first octat ad bouded b clider + 4 ad the plae + z 4. Aswer:. 8

25 Multiple Itegral.4 Chages of variables i Triple Itegrals Let be a ew solid regio, ad u, v, w, u, v, w, z z u, v, w. The V f (,, ) f ( u, v, w), (,, ), z ( u, v, w) du z dv u v w J dvdw V V where J is called Jacobia ad is defied b u v w J u v w z z z u v w.4. From Cartesia to Clidrical Coordiate Sstem z rcosθ rsiθ z z The Jacobia is defied b u v w cosθ r siθ J siθ r cosθ r u v w z z z u v w.4. From Cartesia coordiate to Spherical Coordiate Sstem We have rcosθ, rsiθ ad r ρ siφ, hece we ca fid ρ siφcosθ ρ si φ si θ z ρcosφ ad furthermore we ca obtai the relatio + + z ρ Now we compute the Jacobia. O r θ z φ ρ θ ( r, θ, z) ( r, θ,) ( ρ, θφ, ) ρ θ φ si φ cosθ ρ si φ siθ ρ cosφ cosθ J si φ siθ ρ si φ cosθ ρ cosφ siθ ρ si φ ρ θ φ cosφ ρ si φ z z z ρ θ φ Hece J ρ siφ 9

26 Multiple Itegral Eample 4 Compute circle + dd, where is the regio i -plae, eclosed betwee the + 4 ad + 9. Aswer: 8 Eample 5 Compute dddz where V is a solid above -plae, iside the V clider + a, but below the parabola z +. Aswer: π π 4 a Eample 6 Compute dd where is the regio eclosed i the ellipse +. a b (Hit: let u, v. Aswer: π ab ) a b Eample 7 Use spherical coordiate sstem to compute 4 4 z + + z dzdd Aswer: 4 Applicatios. Computatio of a Plae Area The area of the regio D i is foud b 64π 9 A dd D Eample Fid the area of the regio eclosed betwee π si where 4. Aswer:. cos ad D. The Volume of a Solid (i). The volume of a solid defied b the surfac e z f (,, -plae where (, D (see the figure), f (, ad f is, cotiuous over D, is foud b z z f (, V f (, dd D Eample Fid the volume of a tetrahedro geerated b the three coordiate plaes ad the 4 plae z 4 4. Aswer:. Eample D

27 Multiple Itegral Fid the volume of the solid geerated b z + +, + ad the three coordiate plaes. Aswer: 4 (ii). Le S be the solid regio i the space, the the volume of this solid ca be foud b S Eample 4 Fid the volume of a solid eclosed i sphere + z. Aswer: 9 π 6 V dddz S + + z 4 ad paraboloid f. Surface Area as a Double Itegral Assume that the fuctio f (, has cotiuous partial derivatives ad f i a regio of the - plae. The the portio of the surface z f, that li es over has surface area S ad is foud b S z f (, S [ f (, ] + [ f ( ] + da, Eample 5 Fid the surface area of the portio of the plae + + z that lies i the first octat (where,, z ) Solutio Let z f (,, the f, ad (,. The S + + dd dd f Eample 6 Fid the surface area of that part of the paraboloid π plae z.as: ( 7 ) 6 Eample z 5 that lies above the

28 Multiple Itegral Fid the surface area of the portio of th e clider + z 9 that lies above the { } rectagle, :,. Aswer: 6π.4 Mass ad Ceter of Mass.4. Mass of a Plaar Lamia of Variable Desit A plaar lamia is a flat plate that occupies a regio i the plae ad is so thi that it ca be regarded as two dimesioal. If δ is a cotiuous desit fuctio o the lamia correspodig to a plae regio, the the mass m of the lamia is give b m δ (, da Eample 8 Fid the mass of the lamia of desit δ (, that occupies the regio bouded b the parabola ad the lie Solutio We fid the domai of itegral. B substitutio we have The, Thus, 6 m da dd Eample 9 A triagle lamia with vertices Fid its total mass. Aswer: 4 (, ),(, ),(, ) (, has desit fuctio δ..4. Momet ad Ceter of Mass The momet of a object about a ais measures the tedec of the object to rotate about that ais. It is defied as the product of the object's mass ad the siged distace from the ais. If δ (, is a cotiuous desit fuctio o a lamia correspodig to a plae regio, the the momets of mass with respect to the - ais ad -ais, respectivel, defied b M δ (, da a d M δ (, da (, Furthermore, if m is the mass of the lamia, the ceter of mass is, where M M ad m m, is called the cetroid of the regio. If the desit δ is costat, the poit Eample

29 Multiple Itegral Locate the ceter of mass of the lamia of desit δ (, Solutio regio bouded b the parabola We have M M ad lie 9 (, da dd 7 δ (, δ da 8 dd 5 6 From previous eample we foud mass m. The, M M 7 m 6 7 m Hece the ceter of mass is, that occupies the W e ca use the triple itegral to fid the mass ad ceter of mass of a solid i with desitδ (,, z). The mass m, momets M z, M z, M about the z, z, ad plas, respectivel, ad coordiates,, z of the ceter of mass are give b: Mass m δ (,, z)dv Momets M δ (,, z)dv, is the distace to the z-plae z δ (,, z)dv, is the distace to the z-plae M z (,, z)dv, z is the distace to the -plae M zδ z Ceter of mass (,, z),, m m m M Eample A solid tetrahedro has vertices (,,), (,,) desit δ 6. Fid the cetroid. Solutio The tetrahedro ca be described as the reg beeath the plae + + z. The m δ dv The we fid that z 6dzdd M M, (,, ) ad (,,) ad costat io i the first octat that lies

30 Multiple Itegral M z 6dV dzdd 6 4 Mz 6dV 6 dzdd 4 M 6zdV 6zdzdd 4 M z M M z Thus,,, z m 4 m 4 m 4.5 Momets of Iertia I geeral, a lamia of desit δ (, coverig regio i the first quadrat of the plae has first momet about a lie L give b the itegral M L sdm where d m δ (, daad s s(, is the distace from a tpical poit P, i to L. Similarl, the secod momet of momet of iertia of about L is defied b I sdm L. I phsics, the momets of iertia measure the tedec of the lamia to resist a chage i rotatioal motio about ais L. The momets of iertia of a lamia of desit δ coverig the plae regio about -, -, z-ais, respectivel, are give b z δ (, ) I da I δ (, da z δ, s + I + da I I s s + (, Eample A lamia occupies the regio i the plae that is bouded b the parabola ad the lie s, ad, is (,. The desit of the lamia at each poit δ. Fid the momets of iertia of the lamia about the -ais ad -ais. Solutio dd (, ) I dm δ da dd 56 4 I dm da dd

31 Multiple Itegral Eample A lamia with desit δ, is bouded b the -ais, the lie 8 ad the curve. Fid the momet of iertia about the aes. Solutio I da dd 7 8 I da dd 644 I I + I z We ca calculate the momets of iertia of the solid i follows: I + z δ,, z dv ( ) δ (,, ) ( ) δ (,, ) I + z z dv I + z dv z which are defied as Eample 4 Fid the momet of iertia about the z-ais of the solid tetrahedro S with vertices (,, ),(,, ),(,, ),(,, )ad desit δ (, z, ) Solutio I + δ,, z dv z 9 + dzdd Eercises Let {(, : 4, }. Evaluate f (, ) dawhere, <, a. f (,, 4, (Aswer: 4), <, < b. f (,, <,, 4, (Aswer: ) Evaluate the followig itegral e dd a. dd (Aswer: ) b. (Aswer: e ) 5

32 Multiple Itegral π c. si dd (Aswer: ) d. dd (Aswer: 4 ) e. ( dd(aswer: ) f. dd (Aswer: l ) g. ( + ) ( + ) l l dd(aswer: 8 + ) h. (Aswer: ) e dd l i. dd (Aswer: l ) j. e dd (As: ( l) ) Evaluate the double itegral over the rectagular regio. a. da, :, (Aswer: ) 4 { } b. cos( + da {(, : π 4 π 4, π 4} c. 4 da {, :, } ( (Aswer: ) Evaluate the followig itegrals a. dd (Aswer: 9 4 ) b. c. π π si dd (Aswer: ) d. π a a (Aswer: ) dd (Aswer: 9) π π cos dd (Aswer: ) a e. ( + dd(aswer: ) f. dd 5 6dA, where is the regio bouded b,, ad (As: ) 6 cos da, where is the regio eclosed b 7,, π, ad π. (Aswer: ) π da where is the regio bouded b 6, ad 8. (As: 576) + da, where is the regio i the first quadrat, eclosed b ad. (Aswer: 7 ) da, where is a triagular regio with vertices (, ),(,) ad (,). + π (Aswer: l ) 4 6

33 Multiple Itegral Compute ( ) da, is the regio eclosed betwee ad. (Aswer: ) π si r cosθ drd θ (Aswer: 6 ) π asiθ π rdrdθ (Aswer: ) 4 ( ) da, where eclosed b the circle + (Aswer: π ( e )) e + dawhere is the sector i the first quadrat that is bouded b + + π, ad + 4. (Aswer: l 5 ). 8 π + dd (Asw er : ) a a 6 + dd(aswer: ) 9 dd,( a > ) (Aswer: + + π + a π 4 ( 5 ) 4 dd (Aswer: e dd (Aswer: ( e 6 )) 4 8 si da, is the regio bouded b,, cos8 ) (Aswer: ( + + ) z dddz (Aswer: 8) z zddzd (Aswer: 7) 9 z 8 d ddz (Aswer: 5 ) dzdd 5 ) (Aswer: Use spherical coordiate to compute π (Aswer: ( e ) ) 6 Use spherical coordiate to compute (Aswer:8π ) e dzdd + + z z dzdd 7

34 Multiple Itegral 7 Use double itegral to fid the volume of the solid tetrahedro that lies i the first octat that is bouded b the three coordiate plaes ad the plae z 5. 5 (Aswer: ) 8 Fid the vo lume of the solid that is bouded above b the plae z + + below b the -plae ad laterall b ad. (Aswer: 56 5 ) 9 Use double itegral to fid the volume of the solid that is bouded above b the paraboloid z 9 +, below b the plae z ad laterall b the plaes z,, ad. (Aswer: 7) Use double itegral to fid the volume of the wedge cut from the clider b the pla e ad. (Aswer: 7 π z z + ) Use double itegral i the first octat bouded b the three coord iate plaes ad the plae + 4 ad + 8 4z (Aswer: ) Fid the volume of the solid bouded above b the paraboliod z ad π b elow b the -plae. (Aswer: ) Use double itegral to fid the volume of the solid commo to the cliders + 5 ad + z 5. (Aswer: ) 4 Fid the volume of the solid eclosed b the sphere + + z 9 ad the 4π clider +. (Aswer: ( 7 8 ) 5 Volume of the solid that is bouded above b the coe z +, below b the -plae, ad laterall b the clider +. (Aswer: 9 ) + 6 The itegral e dwhich arises i probabilit theor, ca be evaluated usig a trick. Let the value of the itegral be I. Thus + + I e d e d sice the letter used for the variable of itegratio i a defiite itegral does ot m atter, + + ( + ) a. Show that I e dd b. Evaluate I b coverti g to polar coordiate ad fid I. 7 Fid the surface area of the portio of the paraboloid z + below the π plae z. (Aswer: ( 5 5+ ) ) 6 8 Fid the surface area of the portio of the coe z that is above the regio i the first quadrat bo uded b the lie ad parabola. (Aswer: 5 6 ) 8

35 Multiple Itegral 9 Fid the surface area of the portio of the paraboloid z that is above π the -plae. (Aswer: ( 5 5+ ) ) 6 4 Fid the surface area of the portio of the surface z that is above the sector i the first quadrat bouded b the lie, ad the circle + 9. π (Aswer: ( ) 8 4 Fid the surface area of the portio of the sphere + + z 6 betwee the plae z ad z. (Aswer:8π ) 4 Fid the surface area of the portio of + z 6 that lies iside the circular clider + 6. (Aswer: 8) 4 Compute si zdv, G is the rectagular bo defied b the G iequalities π,, z π 6. (Aswer: ( π ) π ) 44 Use triple itegral to fi d the volume of the solid i the first octat bouded b the coordiate plaes ad the plae+ 6+ 4z. (Aswer: 4). 45 Use triple itegral to fid the volume of the solid bouded b the surface ad plaes + z 4ad z. ( Aswer: 56 5 ). 46 Use triple itegral to fid the volume of the solid eclosed betwee the elliptic clider + 9 9ad the plaes z ad z +. (Aswer:9π ) 47 Use triple itegral to fid the volume of the solid bouded b the paraboloid z 4 + ad parabolic clider z 4. (Asw er: π ). 48 Use triple itegral to fid the volume of the solid that is eclosed betwee the sphere + + z a ad the paraboloid az +. (Aswer: π a ). 49 Let G be the tetrahedro i the first octat bouded b the coordiate plaes ad z the plaes + +, ( a>, b>, c> ). a b c a. List si differet iterated itegrals that represet the volume of G. b. Evaluate a oe of the si to show that the volume of G. (Aswer: abc ). 6 a with desit δ + is bou ded b the -ais, the lie ad the curve. Fid its mass ad ceter of m ass. ( m,(, 9, 6 ) 7 st 5 A lamia with desit δ, is i the quadrat ad is bouded b the 5 A lami (, ) circle + a ad coordiates aes. Fid the mass ad ceter of mass. 4 a 8a 8a ( m,(,, A triagular lamia is bouded b ad, ad -ais. Its desit isδ. Fid cetroid of the lamia. ((, ), ) 9

36 Multiple Itegral 5 A lamia of desit occupies the regio above -ais ad betwee the circles + a ad + b ( a< b). Aswer: ( ) ( b a ) m π b a 4 b a (,, π 54 A cube is defied b the three iequalities a, a, z a, has desit (,, ) 4 a δ z a. Fid its mass ad ceter of mass. Aswer: m ( z) a a a,,,,.

37 Ordiar Differetial Equatio I Chapter Ordiar Differetial Equatio I Itroductio. What is a differetial Equatio? A differetial equatio is a equatio, which cotais derivatives, either ordiar derivatives or partial derivatives. If the ukow fuctio depeds o a sigle real variable, the differetial equatio is called a ordiar differetial equatio. The followigs are the ordiar differetial equatios. d d +, d d, d d + + ( 4) d d I the differetial equatios, the ukow quatit ( ) is called the depedet variable, ad the real variable,, is called the idepedet variable. d d d d I here we defie,,,, d d d d. Order of a Differetial Equatio The order of a differetial equatio is the order of the highest derivative that occurs i the equatio. For eample, d d st order d d + d order d d 4 d 4 d 4th order Defiitio: Ordiar Differetial Equatio A th -order ordiar differetial equatio is a equatio that has the geeral form F,, ', ",..., where the primes deote differetiatio with respect to, that is, d d,, ad so o d d. Liear ad Noliear Differetial Equatios A liear differetial equatio is a differetial equatio that ca be writte i the

38 Ordiar Differetial Equatio I d d d form a ( ) + a ( ) + + a ( ) + a ( ) f ( ) with d d d ot idetical zero. The a are kow fuctios of called coefficiets. A a ( ) i equatio that is ot liear is called oliear. Whe the coefficiets are costat fuctios, the differetial equatio is said to have costat coefficiets. Furthermore, the differetial equatio is said to be homogeeous if f ( ) ad o-homogeeous if f ( ) is ot ideticall zero. Eamples of classificatio of Differetial Equatios: Differetial equatio Liear or Noliear Order Homogeeous or ohomogeeous Costat or variable coefficiets d + d Liear No-homogeeous Variable d d + + d d Noliear No-homogeeous Variable d d + + d d Liear Homogeeous Costat 4 d si 4 d + Liear 4 No-homogeeous Costat.4 Solutios A fuctio is a solutio of a differetial equatio o a iterval if, whe substituted ito the differetial equatio, the resultig equalit is true for all values of i the domai of ( ). Eample Verif that ( ) si + is a solutio of the secod order liear equatio + + Eample Verif that the fuctio ( ) e d is a solutio of the differetial equatio d for all..5 Implicit/Eplicit Solutio A eplicit solutio is a solutio that is give i the form ( ). I some occasios, it is impossible to deduce a eplicit represetatio for i term of. Such solutios are called implicit solutios. Eample

39 Ordiar Differetial Equatio I The relatio e + implicitl defies as a fuctio of. Verif that this implicitl defied fuctio is a solutio of the differetial equatio d d + Solutio Differetiatig e + with respect to gives d d e + d d d ( e + ) d Thus d d e + Substitute this ad e + ito the equatio gives e + e + + or e + e + which is true..6 Iitial-Value Problem (IVP) A iitial-value problem for a th-order equatio d d d F,,,,, d d d cosists i fidig the solutio to the differetial equatio o a iterval I that also ( satisfies the iitial coditios ( ), ( ),, ) ( ) where ad,,, are give costats. Eample 4 Verif that si cos Solutio + is a solutio of the iitial value problem +,, We have ( ) cos si ( ) si cos Substitutig ito the equatio gives si cos si cos + ( ) + ( + ) Hece equatio. To verif that ( ) also satisfies the iitial coditios, we observe that I satisfies the differetial

40 Ordiar Differetial Equatio I si+ cos, cos si.7 Geeral Solutio of a Differetial Equatio I geeral case whe solvig a th-order equatio obtai -parameter famil of solutios (,, ) ( ) F,,,, we geerall G c, c,, c. A solutio of a differetial equatio that is free of arbitrar parameters is called a specific or particular solutio. Eample 5 c The fuctio ( ) + is the geeral solutio to + 4. From this eample 4 9 the fuctio ( ) is the particular solutio whe applig the iitial 4 4 coditio () 4o the equatio + 4 ; that is, it is the solutio to the iitial value problem + 4, 4. () Eercises Show that each fuctio is a solutio of the give differetial equatio. Assume that a ad c are costats.. d a d a e d. e d + e d. a d + csi a d d d d Show that the followig relatio defies a implicit solutio of the give differetial equatio 5. ' e e 6. ' + c Verif that the specified fuctio is a solutio of the give iitial-value problem Differetial Equatio Iitial Coditio(s) Fuctio. ' + e ( ) ( ) + ( ) ' ( ) ( ) cos + + ( ) ' ( ) ( ) e e. '. " 4 5. " ' 4

41 Ordiar Differetial Equatio I Separable Equatios A differetial equatio d f (, ) d is called separable if it ca be writte as d h( ) g( d That is, f ( factors, ) ito a fuctio of times a fuctio of. Either ma be costat so that ever differetial equatio of the form d h( ) d or d g ( ) d is separable. Some eamples of such fuctios are + e e e, ( + )( + ) h or g (Here h( ) ) If f (, ) has bee factored so that the differetial equatio is writte as i the above eamples, the we divide b g( to get d h( ) g( d Net we ati-differetiate both sides with respect to d d h d g d d B the chai rule d d so d h( ) d d g Solutio b Separatio of Variables d To solve f (, d b separatios of variables, we proceed b the followig: (i). Factor f (, ) hg ( (ii). ewrite d d h( ) g ( i differetial form as d h( ) d g( (iii). The solutio is d h( ) d g Eample Solve the differetial equatio Solutio d d + 5

42 Ordiar Differetial Equatio I ewrite the differetial equatio i differetial form as d d + So that d d + The Arc ta + C Hece Eample ta ( + C) d, d Solve the iitial-value problem I differetial form we obtai So d d d d + + C C B substitute the iitial coditio, ito this equatio givesc. Hece the implicit solutio is +. Eample d + + Solve the differetial equatio 6 d + + Eercises for sectio Solve the give differetial equatios d. d 6. d d d d. 7. d d + d +. e d d 8. 5 d + d ( + ) 4. d 9., d ( ) d + d 5. d 6

43 Ordiar Differetial Equatio I d + d., ()., ( ) d 4 d d., () d e du t + 4. dt u + 4 d. 4 d First Order Liear Equatios The first order liear equatio is of the form d a( ) a( ) h( ) d + with a. The equatio ca be rewritte as d p d + p a a g h a. g( ) where ad We ow solve the later equatio. We will tr to fid a fuctio u( ), called a itegratig factor, such that d u( ) + p( ) d To fid u, we proceed as follows: u + u p u + u If we assume that ( ), we arrive at u ( ) p( ) u( ) We ca fid a solutio u( ) > u ( ) p ( ) u( ) l u( ) p( ) d p( ) d u( ) e d u + p d we have but the d( u d b separatig variables, gettig d( u d ( + ) u p u g ( + ) or u u g u u g d + C u p u 7

44 Ordiar Differetial Equatio I Summar of First Order Liear Procedure d + p g d pd b. Compute the itegratig factor u e c. Multipl both sides b u( ) to get d ( u ) ug d d. Ati-differetiate both sides with respect to, u ug d + C a. ewrite the differetial equatio as e. If there is iitial coditio, use it to fid C. f. Solve for. Eample Fid all solutios of ', > Solutio The equatio ca be rewritte as ' the p( ). Thus d u e e e l l Multiplig each side of the differetial equatio b the itegratig factor, we get ' d ( ( )) d + C Eample + C d si +, >, π d Solve the iitial value problem ( ) Solutio Sice p, the itegratig factor is d l μ e e Multiplig both sides of equatios b itegral factor give ' + si d ( si d b itegratio we obtai 8

45 Ordiar Differetial Equatio I cos + C Thus C cos π C Sice,, the, that is C π π 8 π 8 ( > ) cos Hece the solutio is ( ),( > ) 4 Beroulli Equatios A Beroulli Equatios is a first-order differetial equatio i the form d p( ) q( ) d + () If or, the the Beroulli equatio is alread first order liear ad ca be solved b the method of the previous sectio. If ad the the substitutio v will chage the Beroulli equatio to a liear equatio i v ad. Let v, the v v v ( ) or ( ) or d dv d d Substitutig this ito () ields dv + p( ) q( ) d ad use v, we obtai B dividig both sides b dv ( ) p( ) v ( ) q( d + ) which is a liear first-order differetial equatio i v. Eample d Solve the differetial equatio d + Solutio The equatio ca be rewritte as d d 9

46 Ordiar Differetial Equatio I with Let, p ad q v, the ( ) v 4 So v B the substitutio of ito the equatio, we obtai v dv d Dividig both sides b gives dv + d Substitute v i the equatio we obtai dv v d + d We have p( ) the u( ) e e Multiplicatio of the equatio b the itegral factor gives e v + v e ( ve ) e Ati-differetiatig with respect to gives ve e + C Solvig for v, we obtai v + Ce Ce ± Ce Eercises for sectio ad 4 d d si d d + d d. e 5. + d + d + e. d d e 6. d d + 4

47 Ordiar Differetial Equatio I 7. d d + 8. d + d 9. d d + d. cos si d + d. cos d d. ( + e ) + e d Solve the iitial-value problem d. d ( ) d. d + () d. d () 4 d. d + ( ) 5 iccati Equatio The oliear differetial equatio d. ( + 9) + d d e d 5. ' ' + ' + e 8. ' ' + d + + d ' + si, π 4. ( e ) e 5. ' +, 4 ', d f ( ) g( ) h( ) d + + () is called iccati equatio. I order to solve a iccati equatio, oe will eed a particular, the the substitutio solutio. Let coverts the equatio to dw ( ) + w g h w h d which is a liear differetial equatio of first order with respect to the fuctio w w( ) Proof We have ( ) +. Differetiatig with respect to, ields w Substitutig ad d ito (), ields d d d dw d d w d. 4

48 Ordiar Differetial Equatio I d dw f ( ) g ( ) h ( ) + d w d w w d dw f ( ) + g ( ) + g( ) + h( ) + h( ) + h ( ) d w d w w w dw d + f ( ) + g ( ) + g( ) + h( ) + h( ) + h ( ) w d d w w w dw d + f ( ) + g ( ) + g( ) + h( ) + h( ) + h ( ) w d d w w w d Sice f ( ) + g( ) + h( ), we obtai d dw d d + + g ( ) + h( ) + h ( ) w d d d w w w dw + g ( ) + h ( ) + h ( ) w d w w w dw g( ) h( ) w h( ) d Eample d Solve the iccati Equatio d +, give that is a particular solutio. Solutio Substitutig + coverts the equatio to w dw w d + which is a first order liear equatio. Its itegratig factor is d μ e e Multiplig both sides of the equatio b itegratig factor, ields dw e + w e d e w e d e + c w + ce Fiall the geeral solutio to the equatio is 6 Eact Equatios + +ce A differetial equatio (, ) (, ) M d+ N d 4

49 Ordiar Differetial Equatio I is eact if there eists a fuctio F(, such that (, ) (, ) + (, ) df M d N d If M ( ad, ) N(, are cotiuous fuctios ad have cotiuous first partial derivatives o some domai, the the equatio is eact if ad ol if M N To solve the eact equatio, first solve the equatios F M ad F N for F(,. The solutio is give implicitl b F C, where C represets a arbitrar costat. (, ) Method for Solutio of Eact Equatios a. Write the differetial equatio i the form M (, d+ N(, d b. Compute M ad N. If M N, the equatio is ot eact ad this techique will ot work. If M N, the equatio is eact ad this techique will work. c. Either ati-differetiate F M with respect to or F N with respect to. Ati-differetiatig will itroduce a arbitrar fuctio of the other variable. d. Take the result for F from step c. ad substitute for F to fid the arbitrar fuctio. e. The solutio is F(, C. Eample Solve ( + + d+ ( + 4 ) d Solutio I this problem, M + + ad N + 4. Sice M N, the equatio is eact. The F M + +, F M + 4 Either equatio ca be ati-differetiated. We shall ati-differetiate the secod oe: 4 F Fd ( + 4 ) d + + k ( ), where k( ) is the ukow fuctio of. We the substitute this epressio for F i the other equatio F M + + i other to fid k. The or Thus k( ) + k k ( ) + ad 4 (, ) k + F ad the geeral solutio is C 4

50 Ordiar Differetial Equatio I Eample + e Solve e (The solutio is F, + e C ) Some o-eact equatios ca be made eact b the followig procedure. Itegratig Factor Method M, d+ N, d, first compute M ad N For differetial equatio M N N caot be epressed as a fuctio of ol, the we do ot have a itegratig factor that is a fuctio of ol. If Q ( M N) N Q( ) is a fuctio of, the u( ) e dis a itegratig factor. N M M caot be epressed as a fuctio of ol, the we do ot have a itegratig factor that is a fuctio of ol. If ( M N) N ( is a fuctio of, theu( e d is a itegratig factor. M, d+ N, d b itegratig factor a. If ( ) b. If. Multipl. Solve the eact equatio (, ) (, ) (, ) (, ) um d+ u N d Eample Solve the differetial equatio( + 4) d + ( d Solutio I this eample, M + 4 N so M 6 N, ad the equatio is ot eact. How ever M N 6 N is a fuctio of. Thus there is a itegratig factor d ( l ) u( ) e e Multiplig the differetial equatio b gives the ew differetial equatio + 4 d+ d which is eact. The solutio of this equatio ca be foud to be Eercises Solve the differetial equatio + d+ + + d, 5.. ( e ) d+ e d, ( ) 4 F + C 44

51 Ordiar Differetial Equatio I si + cos d+ si + d t tdt+ t t d 5. ( d + ( + d 6. d+ + 4 d 7. + si d+ cos d 8. ( ) d ( cos d, ( ) 9. ( si + e ) d+ ( cos d. d + ( l + ) d Fid a appropriate itegratig factor for each differetial equatio ad the solve + d d d+ d.. d 6. + d 7. d d + + d d 8. d + d 9. d + d + d+ d. ( ) 4. ( ) 5. ( 4 ) + d+ d 7 Homogeeous Equatios I this sectio we develop a substitutio techique that ca sometimes be used whe other techiques fail. Suppose that we have the differetial equatio ad the value of (, ) f ( as, ) a fuctio F of d f (, ) d () f depeds ol o the ratio v, so that we ca thik of, (, ) f F Fv Eamples of such fuctios are: + + /. + + /. F( v) v e F( v) e + + ( ) F ( v) + ( ) + ( ) + v + v + v /. v+ v Usig F, we ca rewrite () as d F () d A differetial equatio i the form () that ma be writte i the form () is sometimes called homogeous. Let v so that () becomes 45

52 Ordiar Differetial Equatio I dv f ( v) or v+ F( v) d v d d which ma alwas be solved b separatio of variables (separable equatio) dv d F v v There is a alterative defiitio of homogeeous that is easier to verif: d f (, ) d is homogeeous equatio if f ( t, t f (, () for all t such that (, ad ( t, t are i the domai of f. Defiitio A fuctio (, ) f ( k, k k f (, f is said to be homogeeous of degree i ad if, for ever k, where k is a real parameter. Eample f, + is homogeeous of degree sice (i). (, ) + ( + ) (, ) f k k k k k k k f (ii). f (, ) e (, ) k k (iii). f (, + +5 is homogeeous of degree zero sice f k k e k e is ot homogeeous. Summar of Method for Homogeeous Equatios (i). Verif that the equatio is homogeeous. (ii). Let v dv d + v F v to get (iii). Solve the separable equatio (iv). Let v to get the aswer i terms of ad Eample d + 5 Solve the differetial equatio d + Solutio The equatio is homogeeous sice t + 5t + 5 f ( t, t f (, t + t + Let v, the equatio becomes 46

53 Ordiar Differetial Equatio I dv + 5v + 5v v+ d + v + v dv + 5v v d + v dv + v v d + v + v v v+ B separatio of variables + v dv d v v+ 4 dv d v v 4l v l v l + C Takig the epoetial of both sides of the last equatio ields 4l v l v C l e e ( v ) ( v ) 4 C where the absolute values have bee dropped b allowig C to take o egative or positive value. Let v to get ( ) 4 C ( ) Eample Solve the differetial equatio ( t + ( t) t ( + ) + f ( t, t f (,, so the equatio is 4 ( t)( t t ( ) homogeeous. ( v) 4 4 dv + v+ d v 4 dv v + d v Separatig variables ields v d dv 4 v + l 4 ( v + ) l + C 4 v + C 4 4 l l e l 4 4 e v + l C 47

54 Ordiar Differetial Equatio I But v, the 4 e l v + e l + k e 4 v + e 4 v + e 4 C C 4C C 4 4 v k k e +, k Eercises Verif that the differetial equatio is homogeeous ad solve it. d + ' 6. d +. + d + ' 7. d. d d + e + 8. d d + 4. d d + 9. d + 4 d 5. d + d + +. d + d 48

55 Ordiar Differetial Equatio II Chapter Ordiar Differetial Equatios II (Higher-Order) Secod Order Noliear Equatios I geeral, secod-order oliear differetial equatios are hard to solve. This sectio will preset two substitutios which allow us to solve several importat equatios of the form d d f,, d d b solvig two first order equatios. Case: Depedet variable missig Suppose the differetial equatio ivolves ol the idepedet variable ad derivatives of the depedet variable : d d f, d d d dv the let v ad the equatio becomes a first order equatio f ( v), d d i v which ca be solved b usig previous techiques. Eample Solve the iitial value problem " ( '), ( ), '( ) Solutio Note that the depedet variable does t appear eplicitl i the equatio. Let v '. The iitial coditio ' is the v( ) ad the differetial equatio is dv v d which ca be solved b separatio of variable. dv d v or + C v Applig the iitial coditio, we obtai C. The d v or d d d + d l d + C 49

56 Ordiar Differetial Equatio II where C is a ew arbitrar costat. B the iitial coditio ( ) it implies that C ad hece the solutio is + l + Eample ' Solve the differetial equatio ", > Solutio Agai the depedet variable is missig from the equatio. Let v '. The equatio becomes dv v d which ca be cosidered a first order liear equatio. The itegratig factor is d l μ ( ) e e Multiplig the equatio b the itegratig factor to get ' v Ati-differetiate ad solve for v to obtai v C or ' C Ati-differetiate agai to fid : C + C Case : Idepedet Variable Missig d d I this case the equatio is of the form f, ; that is, it ivolves d d ol the depedet variable ad its derivatives. Agai let v d d to get dv f ( v), d I order to reduce this to a equatio i just ad v, observe that, b the chai rule, dv dv d dv v d d d d Thus we ca obtai dv v f (, ) d v which is the first-order equatio i v ad. Eample Solve the iitial-value problem " ( '), ( ), '( ) Solutio Note that the idepedet variable is missig from the equatio. Let 5

57 Ordiar Differetial Equatio II d dv d v, v d d d Whe, the, so v. The iitial value problem becomes v dv v () v v d, Proceed b separatio of variables, assumig that v. dv d v ad ati-differetiate, gettig + C v Applig the iitial coditio, we getc, the v / Thus d d This ca be solved b separatio of variables. d d ad ati-differetiatio, + C The iitial coditio ( ) implies C ad the fial result is + or ( 6) / Eercises Solve the give secod-order differetial equatio d d 6. ( ) +. + d d 7. ( ), ( ), ( ). e. +, ( ), ( ) 8., ( ), ( ) 4. +, ( ), ( ) ( ) ( ), ( ), ( ) +,,. Homogeeous Liear Differetial Equatio. Liear Idepedece A Liear differetial equatio is oe of the geeral forms ( ) a + a + + a 5

58 Ordiar Differetial Equatio II ad if a ( ), it is said to be of order. It is called homogeeous if ( ) o-homogeous if ( ) et the o-homogeeous oe. ad. First we focus attetio o the homogeeous case ad Defiitio The fuctio ( ), ( ),, ( ) are said to be liear idepedet if the equatio C + C + + C for costats C,, C has ol the trivial solutio C C C for all i the iterval I. Otherwise the are said to be liearl depedet. Eample The fuctio cos ad are liearl idepedet for the ol oe wa we ca have C cos C + for all is for C ad C both to be. However, the fuctios, si ad cos are liear depedet, because () ( si ) ( cos ) C + C + C solutio. Eample The fuctio e, 4e 4 + 4e + 4e. for C, C, C, which is ot the trivial are liearl depedet o the iterval (, ) + sice. Wroskia Suppose the coefficiets a,, a are cotiuous fuctios of o the iterval a bad,,, are solutios of the homogeeous liear differetial equatio ( ) a + a + + a the the fuctio,,, the determiat are liearl idepedet o [, ] W ( ) ( ) ( ) ab if ad ol if for some o [ ab, ]. The determiat is called the Wroskia fuctio of the fuctios o[ ab, ]. Eample The fuctios e, e, e are solutios of a certai homogeeous liear differetial equatio with costat coefficiets. Show that these solutios are liear idepedet. Solutio 5

59 Ordiar Differetial Equatio II e e e,, 6 W e e e e e e e ( ) e e 9e Hece the fuctios are liearl idepedet. Eample 4 The fuctios si ad cos are solutios of the secod-order equatio + 4. Show that the form a liearl idepedet set of fuctios. Theorem If a,, a are cotiuous fuctios of if a ( ) o the iterval [ ab, ], the the th order homogeeous liear differetial equatio ( ) a + a + + a o [, ], has liearl idepedet solutios, ab ad, b the proper choice of costats c,, c ever solutio of the equatio ca be epressed as c + c + + c Eample 5 (a) Show that e ad e are solutios of + 6 (b) Show that e + 5e is also a solutio of this equatio. (c) Show that e is ot a solutio. eductio of Order Theorem If is a otrivial solutio of the th order homogeeous liear differetial equatio ( ) a( ) + a ( ) + + a( ) the the substitutio v, followed b theq substitutio w v, reduced the equatio to (-) th order equatio. Eample (a) Show that e is a solutio of + +. (b) Use the method of reductio of order to fid a secod liearl idepedet solutio of this differetial equatio ad write the geeral solutio. Solutio (a) Substitutig e, e, e ito the give equatio ields e + e + e which shows that e is a solutio of the give equatiol. (b) Usig the method of reductio of orderw, we let ve,which differetiated twicew, ields ve ve v e ve + ve Substitutig ito the give differetial equatio, we get ( v e v e + ve ) + ( v e ve ) + ve Epadig ad collectig terms ields 5